a) $Fe + H_2SO_4 \to FeSO_4 + H_2$

b) Theo PTHH : $n_{H_2SO_4} = n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)$
$C_{M_{H_2SO_4}} = \dfrac{0,15}{0,05} = 3M$

Sửa đề : 11.2 g sắt

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(0.2....0.2.................0.2\)

\(m_{FeSO_4}=0.2\cdot152=30.4\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.2}{0.05}=4\left(M\right)\)

\(m_{Fe}=\dfrac{11}{56}=0.19\left(mol\right)\)

\(Fe+H_2SO_4->FeSO_4+H_2\)

0.19   0.19            0.19              (mol)

\(m_{FeSO_4}=0.19\cdot152=28.28\left(g\right)\)

\(Cm=\dfrac{0.19}{0.05}=3.8M\)

nFe=11.256=0.2(mol)

mFeSO4=0,2⋅152=30,4(g)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\\ pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\) 
  \(LTL:\dfrac{0,2}{1}< \dfrac{0,25}{1}\) 
=> H₂SO₄ dư 
\(n_{H_2}=n_{H_2SO_4\left(p\text{ư}\right)}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2}=0,2.22,4=4,48l\\ m_{H_2SO_4\left(d\right)}=\left(0,25-0,2\right).98=4,9g\) 
 

c

a,\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:     0,1       0,1          0,1        0,1

b,\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c,\(C_{M_{ddH_2SO_4}}=\dfrac{0,1}{0,2}=0,5M\)

d,\(C_{M_{ddFeSO_4}}=\dfrac{0,1}{0,2}=0,5M\)

ai giúp mình với

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ a.Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b.0,1.......0,1..........0,1...........0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c.C_{MddH_2SO_4}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\ d.C_{MddFeSO_4}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(0.25....0.25.......................0.25\)

\(m_{Fe}=0.25\cdot56=14\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.25}{0.1}=2.5\left(M\right)\)

a. PTHH: Fe+H2SO4-->FeSO4+H2

b. Có nH2=\(\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

Theo pthh trên, nFe=nH2=0.25mol

=> mFe=0.25*56=14g

c. Theo pthh trên, nH2=nH2SO4=0.25mol

Đổi 100ml=0.1l

=> \(C_M=\dfrac{0.25}{0.1}=2.5M\)

\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100\cdot20\%}{98}=\dfrac{10}{49}\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(1...........1\)

\(0.02..........\dfrac{10}{49}\)

\(LTL:\dfrac{0.02}{1}< \dfrac{10}{49}\Rightarrow H_2SO_4dư\)

\(m_{\text{dung dịch sau phản ứng}}=1.6+100=101.6\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{101.6}\cdot100\%=3.15\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(\dfrac{10}{49}-0.02\right)\cdot98}{101.6}\cdot100\%=17.75\%\)

Giúp mình với 

nK2SO3=15.8\158=0.1(mol)

K2SO3+H2SO4→K2SO4+SO2+H2O

0.1...........................0.1..........0.1

VSO2=0.1⋅22.4=2.24(l)

CMK2SO4=0.1\0.2=0.5(M)